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Section 8-6 Testing a Claim About a Standard Deviation or Variance Slide 1 Key Concept This section introduces methods for testing a claim made about a population standard deviation σ or population variance σ 2. The methods of this section use the chi-square distribution that was first introduced in Section 7-5. Slide 2 Requirements for Testing Claims About or 2 1. The sample is a simple random sample. 2. The population has a normal distribution. (This is a much stricter requirement than the requirement of a normal distribution when testing claims about means.) Slide 3 Chi-Square Distribution Test Statistic 2= n (n – 1) s 2 2 = sample size s 2 = sample variance 2 = population variance (given in null hypothesis) Slide 4 P-Values and Critical Values for Chi-Square Distribution Use Table A-4. The degrees of freedom = n –1. Slide 5 Properties of Chi-Square Distribution All values of 2 are nonnegative, and the distribution is not symmetric (see Figure 8-13, following). There is a different distribution for each number of degrees of freedom (see Figure 8-14, following). The critical values are found in Table A-4 using n – 1 degrees of freedom. Slide 6 Properties of Chi-Square Distribution - cont Properties of the Chi-Square Distribution Chi-Square Distribution for 10 and 20 Degrees of Freedom There is a different distribution for each number of degrees of freedom. Figure 8-13 Figure 8-14 Slide 7 Example: For a simple random sample of adults, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A simple random sample of 13 statistics professors yields a standard deviation of s = 7.2. Assume that IQ scores of statistics professors are normally distributed and use a 0.05 significance level to test the claim that = 15. H0: = 15 H1: 15 = 0.05 n = 13 s = 7.2 = 2 (n – 1)s2 2 2 (13 – 1)(7.2) = = 2.765 152 Slide 8 Example: For a simple random sample of adults, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A simple random sample of 13 statistics professors yields a standard deviation of s = 7.2. Assume that IQ scores of statistics professors are normally distributed and use a 0.05 significance level to test the claim that = 15. H0: = 15 H1: 15 = 0.05 n = 13 s = 7.2 2 = 2.765 Slide 9 Example: For a simple random sample of adults, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A simple random sample of 13 statistics professors yields a standard deviation of s = 7.2. Assume that IQ scores of statistics professors are normally distributed and use a 0.05 significance level to test the claim that = 15. H0: = 15 H1: 15 = 0.05 n = 13 s = 7.2 2 = 2.765 The critical values of 4.404 and 23.337 are found in Table A-4, in the 12th row (degrees of freedom = n – 1) in the column corresponding to 0.975 and 0.025. Slide 10 Example: For a simple random sample of adults, IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. A simple random sample of 13 statistics professors yields a standard deviation of s = 7.2. Assume that IQ scores of statistics professors are normally distributed and use a 0.05 significance level to test the claim that = 15. H0: = 15 H1: 15 = 0.05 n = 13 s = 7.2 2 = 2.765 Because the test statistic is in the critical region, we reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the standard deviation is equal to 15. Slide 11 Recap In this section we have discussed: Tests for claims about standard deviation and variance. Test statistic. Chi-square distribution. Critical values. Slide 12

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